I was reading "Maximum Likelihood Estimation for the Negative Binomial Dispersion Parameter" by Walter W. Pieogorsch, and in the intro it says Poisson is a limiting case of negative binomial when dispersion papameter $a$ goes to zero:

$$lim_{a\to0}Pr(Y=y)={ \Gamma(y+a^{-1})\over y!\Gamma(a^{-1})}({au\over 1+au})^y(1+au)^{-1/a} = {\mu^ye^{-u} \over y! }$$

I tried to work out the math, and I can see $y!$ stays the same and that $$lim_{a\to0}(1+au)^{-1/a}=e^{-u}$$

but I cannot see how $$lim_{a\to0}{ \Gamma(y+a^{-1})\over\Gamma(a^{-1})}({au\over 1+au})^y = \mu^y$$ How is this possible?

I was reading "Maximum Likelihood Estimation for the Negative Binomial Dispersion Parameter" by Walter W. Pieogorsch, and in the intro it says the Poisson distribution is a limiting case of negative binomial distribution when the dispersion parameter $a$ goes to zero:

$$lim_{a\to0}Pr(Y=y)={ \Gamma(y+a^{-1})\over y!\Gamma(a^{-1})}({au\over 1+au})^y(1+au)^{-1/a} = {\mu^ye^{-u} \over y! }$$

I tried to work out the math, and I can see $y!$ stays the same and that $$lim_{a\to0}(1+au)^{-1/a}=e^{-u}$$

but I cannot see how $$lim_{a\to0}{ \Gamma(y+a^{-1})\over\Gamma(a^{-1})}({au\over 1+au})^y = \mu^y$$ How is this possible?